A coin is tossed three times in succession. If E is the event that there are at least two heads and F is the event in which first throw is a head, then $P\left(\frac{E}{F}\right)=$

$\frac{3}{4}$

$S= \begin{Bmatrix}HHH, HHT, HTH,, THH, HTT, THT,, TTH,, TTT \end{Bmatrix}$

$n(E)= 4, n(F)= 4 $ and $ n(E ∩ F) = 3$

$∴ P\left(\frac{E}{F}\right) = \frac{P(E ∩ F)}{P(F)}=\frac{3/8}{4/8}=\frac{3}{4}.$